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How do I evaluate the Cauchy Integral Formula $f(A)=\frac{1}{2\pi i}\int\limits_Cf(z)(zI-A)^{-1}dz$ for a matrix $A=\left(\begin{array}{ccc}2&2&-5\\3&7&-15\\1&2&-4\end{array}\right)$ and a function $f(x)=3x^2+1$?

I have evaluated the function directly, using interpolation and using Jordan-Normal Form and want to show the solutions are equivalent.

The solution should be $f(A)=\left(\begin{array}16&24&-60\\36&76&-180\\12&24&-56\end{array}\right)$.

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up vote 3 down vote accepted

Choose the contour $C$ so that $|z| > \|A\|$ for all $z$ in $C$. Then note that $(zI -A)^{-1} = \frac{1}{z}(I-\frac{A}{z})^{-1}$, and for $|z| > \|A\|$, we have $(zI -A)^{-1} = \frac{1}{z} \sum_{k=0}^\infty \frac{A^k}{z^k}$. Since $|z| > \|A\|$ for all $z$ in $C$, the convergence is uniform so we may interchange integration and summation.

Also note that $f$ is analytic on $C$ and the 'inside' of $C$.

This gives \begin{eqnarray} f(A)&=&\frac{1}{2\pi i}\int\limits_Cf(z)(zI-A)^{-1}dz \\ &=& \frac{1}{2\pi i}\int\limits_Cf(z) \frac{1}{z} \sum_{k=0}^\infty \frac{A^k}{z^k} dz \\ &=& \frac{1}{2\pi i} \sum_{k=0}^\infty \left(\int\limits_Cf(z) \frac{1}{z^{k+1}} dz \right) A^k \\ &=& \sum_{k=0}^\infty \left( \frac{1}{2\pi i}\int\limits_Cf(z) \frac{1}{z^{k+1}} dz \right) A^k \\ &=& \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} A^k \end{eqnarray}

(The specific value doesn't matter, but it is easy to compute $\|A\|_1 = 24$, so as long as $|z|> 24$ on $C$, the above formula holds.)

It follows that if $f(x) = \sum_{k=0}^n a_k z^k$, then $f(A) = \sum_{k=0}^n a_k A^k$. Hence, in this case, $f(A) = 3 A^2 +I$. Evaluating shows that it equals the answer above.

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Thanks. If, for example, $f(x)=\sin{x}$, then, if at all, how would the calculation differ? – hello123 Mar 21 '13 at 22:18
Since $\sin$ is entire, the above formula still holds, but evaluating $A^k$ becomes the issue. Cayley Hamilton lets you replace $A^n$ by a linear combination of smaller powers, but whether or not that is a computation simplification depends on the specific $A$. – copper.hat Mar 21 '13 at 22:21
Could you explain why it is that the convergence is uniform? – hello123 Apr 26 '13 at 13:23
The contour $C$ is compact, hence we have $|z| \ge \beta > \|A\|$ for all $z \in C$. Hence $\frac{\|A\|}{|z|} \le \frac{\|A\|}{\beta} <1$ for all $z \in C$. Then $\sum_{k>n} \frac{\|A\|^k}{|z|^k} \le \sum_{k>n} \frac{\|A\|^k}{\beta^k}$ which can be made as small as you like by choosing $n$ sufficiently large. – copper.hat Apr 26 '13 at 14:40

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